Problem: The graph of a parabola has the following properties:

$\bullet$ It passes through the point $(1,5).$

$\bullet$ The $y$-coordinate of the focus is 3.

$\bullet$ Its axis of symmetry is parallel to the $x$-axis.

$\bullet$ Its vertex lies on the $y$-axis.

Express the equation of the parabola in the form
\[ax^2 + bxy + cy^2 + dx + ey + f = 0,\]where $a,$ $b,$ $c,$ $d,$ $e,$ $f$ are integers, $c$ is a positive integer, and $\gcd(|a|,|b|,|c|,|d|,|e|,|f|) = 1.$
Since the axis of symmetry is parallel to the $x$-axis, and the $y$-coordinate of the focus is 3, the $y$-coordinate of the vertex is also 3.  Since the vertex lies on the $y$-axis, it must be at $(0,3).$  Hence, the equation of the parabola is of the form
\[x = k(y - 3)^2.\][asy]
unitsize(1 cm);

real upperparab (real x) {
  return (sqrt(4*x) + 3);
}

real lowerparab (real x) {
  return (-sqrt(4*x) + 3);
}

draw(graph(upperparab,0,2));
draw(graph(lowerparab,0,2));
draw((0,-1)--(0,6));
draw((-1,0)--(3,0));

dot("$(1,5)$", (1,5), NW);
dot("$(0,3)$", (0,3), W);
[/asy]

Since the graph passes through $(1,5),$ we can plug in $x = 1$ and $y = 5,$ to get $1 = 4k,$ so $k = \frac{1}{4}.$

Hence, the equation of the parabola is $x = \frac{1}{4} (y - 3)^2,$ which we write as
\[\boxed{y^2 - 4x - 6y + 9 = 0}.\]